Rigid Body Motion - cylinder on a plane

[Salviati]

1. The problem statement, all variables and given/known data
http://i43.tinypic.com/280qk4l.jpg
Pretty picture:
http://i41.tinypic.com/j5f3ic.jpg

I am currently stuck on #1.

2. Relevant equations
Not exactly sure but...
For #1, I assume the relevant equations are those relating the (i) gravitational force to momentum and (ii)torque due to gravity to angular momentum:

(i) F = dP/dt
(ii) K = dM/dt, where K = r x F


3. The attempt at a solution
I don't know how to begin. I figured maybe if I find r of the center of mass, take the absolute value of the derivative with respect to time then I'll know the speed...but where do I place the coordinate system then? Or I thought, if I use the equations listed above to find the momentum and angular momentum, I could do something with those to find the speed. I just don't know, quite confused right now.

[Salviati]

Oookay, here's a more serious attempt at an answer:

The instantaneous axis of rotation for this problem is the axis where the cylinder is in contact with the plane. So, the motion of the cylinder can be described as a rotation around this axis (not sure how to imagine this, but ok). This simplifies the problem because the distance of the center of mass from this axis is:
\sqrt{a^2+R^2-2aRcos\phi}, where \phi is the angle between the perpendicular from the instantaneous axis of rotation and the center of mass. Therefore, the velocity of the cylinder is:
V = \dot{\phi}\sqrt{a^2+R^2-2aRcos\phi}.

The immediate problem that I see with this answer is that it doesn't incorporate the moment of inertia, I, like it says it should in the question.

[vela]

Oookay, here's a more serious attempt at an answer:

The instantaneous axis of rotation for this problem is the axis where the cylinder is in contact with the plane. So, the motion of the cylinder can be described as a rotation around this axis (not sure how to imagine this, but ok). This simplifies the problem because the distance of the center of mass from this axis is:
\sqrt{a^2+R^2-2aRcos\phi}, where \phi is the angle between the perpendicular from the instantaneous axis of rotation and the center of mass. Therefore, the velocity of the cylinder is:
V = \dot{\phi}\sqrt{a^2+R^2-2aRcos\phi}.

The immediate problem that I see with this answer is that it doesn't incorporate the moment of inertia, I, like it says it should in the question.
Your answer looks fine. Just make sure it gives you sensible results for easy-to-check cases. I'm not sure why the problem mentioned I since it's not a geometrical parameter.